summation problem - Page 2

Sneaky · #16 $Old$ November 7th, 2009, 08:40 PM

ok i have a new problem,

99
E (1/i) - (1/(i+1))
i=3

i simplify it to

E 1
----------
E i^2 + E i

how do i move on, because now i = 3 in the E

VonNemo19 · #17 $Old$ November 7th, 2009, 08:47 PM

Quote:

Originally Posted by Sneaky $View Post$

ok i have a new problem,

99
E (1/i) - (1/(i+1))
i=3

i simplify it to

E 1
----------
E i^2 + E i

how do i move on, because now i = 3 in the E

Sneaky. You don't have do do all of that! Recognize what's gonna cancel, and you will see that there will only be two terms left!

Sneaky · #18 $Old$ November 7th, 2009, 08:50 PM

ok i am lost again, can you show me how to do it and include the answer

VonNemo19 · #19 $Old$ November 7th, 2009, 08:58 PM

Quote:

Originally Posted by Sneaky $View Post$

ok i am lost again, can you show me how to do it and include the answer

Check it out

$\sum_{i=3}^{99}\left(\frac{1}{i}-\frac{1}{i+1}\right)=\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)+\left(\frac{1}{99}-\frac{1}{100}\right)$

So, you can see all of the cancellation that is gonna occur! The only thing that'll be left standing is

$=\frac{1}{3}-\frac{1}{100}$